Hum, from Alg class I don't agree that determining if two reg langs
are equal is NP-hard. I believe we can take any regular expression
and convert it to a NFA in P-time. We can take two NFA's and take
their Unions, Compliments and Intersections in P-time. Then we can
create new NFA that is (L1 intersect L2') union (L1' intersect L2).
If L1 = L2 then L3 is empty. We can determine in P-time a NFA/DFA
is empty. Therefore, isn't the above problem P, and not NP?